Consider a voltaic cell based on these half-cell reactions
$\begin{array}{l} {Ag}_{(aq)}^{+} + e^{-} ⟶ {Ag}_{(s)}; E^{0} = + 0.80 V \\ {Cd}_{(aq)}^{+ 2} + 2 e^{-} ⟶ {Cd}_{(s)}; E^{o} = - 0.40 V \end{array}$
Identify the anode and give the voltage of this cell under standard conditions

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$Ag; E_{cell}^{0} = 2.00 V$

$Ag; E_{cell}^{o} = 0.40 V$

$Cd; E_{cell}^{0} = 2.00 V$

$Cd; E_{cell}^{o} = 1.20 V$

answer is C.

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Detailed Solution

$Cathode : {Ag}^{+} + e^{-} \to Ag {E_{1}}^{o} = + 0.8 V Anode : Cd \to {Cd}^{2 +} + 2 e^{-} {E_{2}}^{o} = + 0.4_________________________{Ag}^{+} + {Cd}^{2 +} \to {Ag}^{+} + Cd E_{3}^{0} = ? △ G_{3} = △ G_{1} + △ G_{2} △ G = - nFE so E_{3}^{o} = \frac{2 \times 0.8 + 2 \times 0.4}{2} = + 1.2 V$