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Question

Calculate the emf of the cell in which of the following reaction takes place:

$$Ni(s) + 2Ag^+ (0.002 M) \rightarrow Ni^{2+} (0.160 M) + 2Ag(s)$$

[Given that $$E^{\circ}_{cell} = 1.05 V$$]

A
$$0.73 V$$
B
$$0.91 V$$
C
$$0.62 V$$
D
$$0.34 V$$
Solution
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Correct option is B. $$0.91 V$$
Given,

$$[Ag^+]=0.002M$$

$$[Ni^{2+}]=0.160M$$

$$n=2$$

According to Nernst equation:

$$E_{cell}=E^0-\dfrac{0.0591}{2}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2}$$

$$\therefore E_{cell}=1.05-\dfrac{0.0591}{2}\log \dfrac{0.160}{(0.002)^2}$$

$$\therefore E_{cell}=1.05-0.02955\, \log{(4\times 10^4)}$$

$$\therefore E_{cell}=0.91V$$

Hence, the correct option is $$\text{B}$$

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[Given that $$E^{\circ}_{cell} = 1.05 V$$]
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