# Predicting the direction of redox reactions  Know that standard electrode potentials can be listed as an electrochemical series.  Use E values to predict.

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Predicting the direction of redox reactions  Know that standard electrode potentials can be listed as an electrochemical series.  Use E values to predict the direction of simple redox reactions and to calculate the e.m.f. of a cell.

GOLDEN RULE The more +ve electrode gains electrons (+ charge attracts electrons)

Electrodes with negative emf are better at releasing electrons (better reducing agents).

–+0 – 0.76 V –ve electrode Zn 2+ + 2 e -  Zn + 0.34 V +ve electrode Cu 2+ + 2 e -  Cu + 1.10 V e–e– Cu 2+ + Zn → Cu + Zn 2+

USE OF Eo VALUES - WILL IT WORK? E° values Can be used to predict the feasibility of redox and cell reactions In theory ANY REDOX REACTION WITH A POSITIVE E° VALUE WILL WORK An equation with a more positive E° value reverse a less positive one

USE OF Eo VALUES - WILL IT WORK? What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equationsCu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V Sn2+(aq) + 2e¯ Sn(s) ; E° = -0.14V the half reaction with the more positive E° value is more likely to work it gets the electrons by reversing the half reaction with the lower E° value therefore Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq) the overall reaction isCu2+(aq) + Sn(s) ——> Sn2+(aq) + Cu(s) the cell voltage is the difference in E° values... (+0.34) - (-0.14) = + 0.48V An equation with a more positive E° value reverse a less positive one

USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous? Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Write out the appropriate equations Cu2+(aq) + 2e¯ Cu(s) ; E° = +0.34V as reductions with their E° values Sn2+(aq) + 2e¯ Sn(s) ; E° = - 0.14V The reaction which takes place will involve the more positive one reversing the other i.e.Cu2+(aq) ——> Cu(s) and Sn(s) ——> Sn2+(aq) The cell voltage will be the difference in E° values and will be positive... (+0.34) - (- 0.14) = + 0.48V If this is the equation you want then it will be spontaneous If it is the opposite equation (going the other way) it will not be spontaneous

USE OF Eo VALUES - WILL IT WORK? An equation with a more positive E° value reverse a less positive one Will this reaction be spontaneous? Sn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Split equation into two half equationsCu2+(aq) + 2e¯ ——> Cu(s) Sn(s) ——> Sn2+(aq) + 2e¯ Find the electrode potentialsCu2+(aq) + 2e¯ Cu(s); E° = +0.34V and the usual equationsSn2+(aq) + 2e¯ Sn(s); E° = - 0.14V Reverse one equation and its signSn(s) ——> Sn2+(aq) + 2e¯ ; E° = +0.14V Combine the two half equationsSn(s) + Cu2+(aq) ——> Sn2+(aq) + Cu(s) Add the two numerical values(+0.34V) + (+ 0.14V) = +0.48V If the value is positive the reaction will be spontaneous

Predicting redox reactions 5.3 exercise 2

–+0 – 0.76 V –ve electrode Zn 2+ + 2 e -  Zn – 0.25 V +ve electrode Ni 2+ + 2 e -  Ni + 0.51 V e–e– Ni 2+ + Zn → Ni + Zn 2+ PREDICTING REDOX REACTIONS – Q1

+ 0 + 0.34 V –ve electrode Cu 2+ + 2 e -  Cu + 0.80 V +ve electrode Ag + + e -  Ag + 0.46 V e–e– 2 Ag + + Cu → 2 Ag + Cu 2+ PREDICTING REDOX REACTIONS – Q2

0 – 2.36 V –ve electrode Mg 2+ + 2 e -  Mg – 0.26 V +ve electrode V 3+ + e -  V 2+ + 2.10 V e–e– Mg(s)|Mg 2+ (aq)||V 3+ (aq),V 2+ (aq)|Pt(s) PREDICTING REDOX REACTIONS – Q3 a – YES: Mg reduces V 3+ to V 2+

0 + 0.77 V –ve electrode Fe 3+ + e -  Fe 2+ + 1.36 V +ve electrode Cl 2 + 2 e -  2 Cl - + 0.59 V e–e– PREDICTING REDOX REACTIONS – Q3 b + NO: Cl - won’t reduce Fe 3+ to Fe 2+

0 + 1.09 V –ve electrode + 1.36 V +ve electrode Cl 2 + 2 e -  2 Cl - + 0.27 V e–e– PREDICTING REDOX REACTIONS – Q3 c + YES: Cl 2 oxidises Br - to Br 2 Br 2 + 2 e -  2 Br - Pt(s)|Br - (aq),Br 2 (aq)||Cl 2 (g)|Cl - (aq)|Pt(s)

0 – 0.14 V –ve electrode Sn 2+ + 2 e -  Sn + 0.77 V +ve electrode Fe 3+ + e -  Fe 2+ + 0.91 V e–e– PREDICTING REDOX REACTIONS – Q3 d – YES: Sn reduces Fe 3+ to Fe 2+ + Sn(s)|Sn 2+ (aq)||Fe 3+ (aq),Fe 2+ (aq)|Pt(s)

0 + 1.33 V –ve electrode Cr 2 O 7 2- + 14 H + + 6 e -  2 Cr 3+ + 7 H 2 O + 1.36 V +ve electrode Cl 2 + 2 e -  2 Cl - + 0.03 V e–e– PREDICTING REDOX REACTIONS – Q3 e + NO: H + /Cr 2 O 7 2- won’t oxidise Cl - to Cl 2

0 + 1.36 V –ve electrode MnO 4 - + 8 H + + 5 e -  Mn 2+ + 4 H 2 O + 1.51 V +ve electrode Cl 2 + 2 e -  2 Cl - + 0.03 V e–e– PREDICTING REDOX REACTIONS – Q3 f + YES: H + /MnO 4 - oxidises Cl - to Cl 2 Pt(s)|Cl - (aq)|Cl 2 (g)||MnO 4 - (aq),H + (aq),Mn 2+ (aq)|Pt(s)

0 – 0.44 V –ve electrode Fe 2+ + 2 e -  Fe 0.00 V +ve electrode 2 H + + 2 e -  H 2 + 0.44 V e–e– PREDICTING REDOX REACTIONS – Q3 g – YES: H + oxidises Fe to Fe 2+ Fe(s)|Fe 2+ (aq)||H + (aq)|H 2 (g)|Pt(s)

0 0.00 V –ve electrode Cu 2+ + 2 e -  Cu + 0.34 V +ve electrode 2 H + + 2 e -  H 2 + 0.34 V e–e– PREDICTING REDOX REACTIONS – Q3 h + NO: H + won’t oxidise Cu to Cu 2+

0 + 1.36 V MnO 4 - + 8 H + + 5 e -  Mn 2+ + 4 H 2 O + 1.51 V Cl 2 + 2 e -  2 Cl - PREDICTING REDOX REACTIONS – Q4 + + 1.33 V Cr 2 O 7 2- + 14 H + + 6 e -  2 Cr 3+ + 7 H 2 O + 0.77 V Fe 3+ + e -  Fe 2+ YES NO

+ 0 ? V –ve electrode Be 2+ + 2 e -  Be + 0.34 V +ve electrode Cu 2+ + 2 e -  Cu + 2.19 V e–e– Be 2+ + Cu → Be + Cu 2+ PREDICTING REDOX REACTIONS – Q5a 2.19 = 0.34 - E  left E  left = 0.34 – 2.19 = – 1.85 V

– 0 ? V –ve electrode Th 4+ + 4 e -  Th + 0.00 V +ve electrode 1.90 V e–e– 4 H + + Th → 2 H 2 + Th 4+ PREDICTING REDOX REACTIONS – Q5b When using SHE E  = cell emf = – 1.90 V 2 H + + 2 e -  H 2

0 0.00 V –ve electrode + 1.09 V +ve electrode + 1.09 V e–e– PREDICTING REDOX REACTIONS – Q6a + Br 2 + 2 e -  2 Br - Pt(s)|H 2 (g)|H + (aq)||Br 2 (aq),Br - (aq)|Pt(s) 2 H + + 2 e -  H 2 H 2 + Br 2 → 2 H + + 2 Br -

0 + 0.34 V –ve electrode + 0.77 V +ve electrode + 0.43 V e–e– PREDICTING REDOX REACTIONS – Q6b + Fe 3+ + e -  Fe 2+ Cu(s)|Cu 2+ (aq)||Fe 3+ (aq),Fe 2+ (aq)|Pt(s) Cu 2+ + 2 e -  Cu 2 Fe 3+ + Cu → 2 Fe 2+ + Cu 2+

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