First slide

Electrochemical cells

Question

$EMF of cell Ni {Ni}^{2 +} (1 .0 M) ∥ {Au}^{3 +} (1 .0 M) Au is......., if E^{\circ} for$ ${Ni}^{2 +} ∣ Ni is - 0 .25 V, E^{\circ} for {Au}^{3 +} ∣ Au is 1 .50 V .$

Moderate

A

+1.255V
B

-1.75 V
C

+1.75 V
D

+4.0 V

Solution

$E_{cell} = E_{{OP}_{Ni / {Ni}_{2}^{2 +}}}^{\circ} + E_{{RP}_{Au}^{3 +}_{/ Au}}^{\circ}$
$\begin{array}{l} E_{{OR}_{Ni}}^{\circ} - \frac{0 .059}{2} \log [{Ni}^{2 +}] + E_{{RP}_{Au}}^{\circ} + \log \frac{0 .059}{3} [{Au}^{3 +}] \\ = 0 .25 - \frac{0 .059}{2} \log (1 .0) + 1 .50 + \frac{0 .059}{3} \log 1 .0 \\ = 1 .75 V \end{array}$

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