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Find the emf of the cell: $\\$ Ni(s) + 2Ag $^+(0.001 M) \rightarrow$ Ni $^{2+} (0.001 M) + 2Ag(s) \\$ (Given that $E^\circ_{cell} = 1.05$ V, $\frac{2.303RT}{F} = 0.059$ at 298 K)

medium

Electrochemistry

2022

chemistry

$1.0385$ V

$1.385$ V

$0.9615$ V

$1.05$ V

Explanation

To find the emf of the cell, we will use the Nernst equation. $\\$ The Nernst equation is given by: $\\ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q \\$

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Find the emf of the cell: \\ Ni(s) + 2Ag+(0.001M)→^+(0.001 M) \rightarrow +(0.001M)→Ni2+(0.001M)+2Ag(s)^{2+} (0.001 M) + 2Ag(s) \\ 2+(0.001M)+2Ag(s)(Given that Ecell∘=1.05 E^\circ_{cell} = 1.05 Ecell∘​=1.05 V, 2.303RTF=0.059 \frac{2.303RT}{F} = 0.059 F2.303RT​=0.059 at 298 K)

Explanation

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Find the emf of the cell: $\\$ Ni(s) + 2Ag $^+(0.001 M) \rightarrow$ Ni $^{2+} (0.001 M) + 2Ag(s) \\$ (Given that $E^\circ_{cell} = 1.05$ V, $\frac{2.303RT}{F} = 0.059$ at 298 K)