Which cell will measure standard electrode potential of copper electrode?

Easy

A

$Pt (s) ∣ H_{2} (g, 0 .1 bar) |H^{+} (aq \cdot 1 M)| {Cu}^{2 +} (aq, 1 M) ∣ Cu$
B

$Pt (s) ∣ H_{2} (g, 1 bar) |H^{+} (aq, 1 M)| {Cu}^{2 +} (aq, 2 M) ∣ Cu$
C

$Pt (s) ∣ H_{2} (g, 1 bar) |H^{†} (aq, 1 M)| {Cu}^{2 +} (aq, 1 M) ∣ Cu$
D

$Pt (s) ∣ H_{2} (g, 0 .1 bar) |H^{+} (aq, 0 .1 M)| {Cu}^{2 +} (aq, 1 M) ∣ Cu$

Solution

Standard electrode potential of copper electrode can be calculated by constructing a concentration cell composed of two half-cell reactions in which concentration of species on left hand and right hand side are unity.
In such case cell potential is equal to standard electrode potential.
$\underset{Oxidation half -cell reaction}{\underset{⏟}{Pt (s) |H_{2} (g, 1 bar)| ∣ H^{+} (aq, 1 M)}} \underset{Reduction half -cell reaction}{\underset{⏟}{|{Cu}^{2 +} (aq, 1 M)| Cu}}$

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