Presentation on theme: "STANDARD ELECTRODE POTENTIALS. THE STANDARD HYDROGEN ELECTRODE In order to measure the potential of an electrode, it is compared to a reference electrode."— Presentation transcript:
THE STANDARD HYDROGEN ELECTRODE In order to measure the potential of an electrode, it is compared to a reference electrode – the standard hydrogen electrode. The standard half reaction is the reduction of hydrogen: 2H + + 2e - → H 2 The electrode potential of this half cell = 0,00V
Measuring the standard electrode potential of zinc by connecting a zinc electrode to the hydrogen electrode.
The hydrogen electrode is the cathode. 2H + (aq) + 2e - → H 2(g) The zinc electrode is the anode. Zn (s) → Zn 2+ (aq) + 2e - The reading on the voltmeter is 0,76V and because electrons flow from the zinc to the hydrogen electrode E o for the zinc half cell = - 0,76V
Full Redox potential table Any substance on the right will spontaneously react with something above it on the right of the table – and vice versa.
When using the table: strong reducing agents are at the top of the table. when combining half reactions, the one higher up the table will be the reducing agent. when combining half reactions, the half reaction located higher up the table is written in reverse (from left to right) and the one lower down from right to left.
Using the table of standard electrode potentials: determine the emf of a cell. predicting whether a redox reaction will occur spontaneously; balancing redox reactions.
Determining the emf of a cell: E o cell = E o reduction – E o oxidation = E o oxidising agent – E o reducing agent = E o cathode – E o anode
For the cell: Zn / Zn 2+ // Cu 2+ / Cu The zinc electrode is the reducing agent and the anode as oxidation occurs at that electrode. E o cell = E o Cu – E o Zn = 0,34 – (-0,76) = +1,10V
Predicting whether a redox reaction will occur spontaneously. Write down the equation as you expect it to occur. from the equation decide which half cell is the reducing/oxidising agent. Determine E o cell based on this information. If E o cell is a positive number, the reaction, as written, is non-spontaneous.
Will the following reaction occur spontaneousky: Pb 2+ (aq) + 2Br - (aq) → Br 2(l) + Pb (s) Pb 2+ + 2e - → PbE o = - 0.13V reduction 2Br - → Br 2 + 2e - E o = + 1,06V oxidation E o cell = E o Pb – E o Br = (- 0,13) – (+ 1,06) = - 1,19V E o is negative and so the reaction is non- spontaneous
Balancing redox equations. Hydrogen sulphide reduces potassium dichromate. Write a balanced equation for this reaction. H 2 S → 2H + + S + 2e - Cr 2 O 7 2- + 14H + + 6e - → Cr 3+ + 7H 2 O Multiply the first equation by 3 to balance the electrons, cancel out any ions/molecules that occur on opposite sides of the equation and add up the remaining ions/molecules to give the balanced ionic equation.
3H 2 S → 6H + + 3S + 6e - Cr 2 O 7 2- + 14H + + 6e - → Cr 3+ + 7H 2 O 3H 2 S + Cr 2 O 7 2- + 8H + → Cr 3+ + 7H 2 O + 3S