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Electrode potential
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+ +1a) Ag(s)/𝐴𝑔 (aq) ‖ 𝐻 (aq)/H2(g):
Step 1 - Write 2 redox half reactions:
The oxidation half reaction is Ag(s) → Ag+ (aq) + e-.
The reduction half reaction is H+ (aq) + 2e- → H2(g).
Step 2 - Find E° reduction:
2H+ (aq) + 2e- → H2(g) E° = +0 V
Step 3 - Find E° oxidation using sign reversal:
This is the reverse of your oxidation reaction:Ag+ (aq) + e- → Ag(s)
Use the potential from the chart to determine E° oxidation: Ag+ (aq) + e- → Ag(s) E° = +0.80 V
To convert from reduction to an oxidation potential, reverse the sign:
E° oxidation = -E° reduction
= - (+0.80 V)
= - 0.80 V
Step 4 - Add the potentials to determine E° cell:
E° cell = E° reduction + E° oxidation
= (+ 0 V) + (- 0.80 V)
= +0 V - 0.80 V
= - 0.80 V
Therefore, the E° cell is - 0.80 V. Since the E° cell is a negative value, the reaction is
nonspontaneous.
Balanced net ionic equation:
+ +
2𝐴𝑔(𝑠) + 2𝐻 (𝑎𝑞) → 2𝐴𝑔 (𝑎𝑞) + 𝐻2(𝑔)
+ 2+
b) H2(g)/𝐻 (aq) ‖ 𝑍𝑛 (aq)/Zn(s):
Step 1 - Write 2 redox half reactions:
The oxidation half reaction is H2(g) → H+ (aq) + 2e-.
2+ −
The reduction half reaction is 𝑍𝑛 (𝑎𝑞) + 2𝑒 → 𝑍𝑛(𝑠).
Step 2 - Find E° reduction:
Zn2+ (aq) + 2e- → Zn(s) E° = -0.76 V
Step 3 - Find E° oxidation using sign reversal:
This is the reverse of your oxidation reaction: 2H+ (aq) + 2e- → H2(g)
Use the potential from the chart to determine E° oxidation: 2H+ (aq) + 2e- → H2(g) E° = + 0 V
To convert from reduction to an oxidation potential, reverse the sign:
E° oxidation = -E° reduction
= - (+ 0 V)
=-0V
Step 4 - Add the potentials to determine E° cell:
E° cell = E° reduction + E° oxidation
= (- 0.76 V) + (- 0 V)
= - 0.76 V -0V
= - 0.76 V
Therefore, the E° cell is - 0.76 V. Since the E° cell is a negative value, the reaction is
nonspontaneous.
Balanced net ionic equation:
2+ +
𝐻2(𝑔) + 𝑍𝑛 (𝑎𝑞) → 2𝐻 (𝑎𝑞) + 𝑍𝑛(𝑠)
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