Calculate the emf of the cell in which the fo | vedclass

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(A) Applying the Nernst equation:$E_{(cell)} = E^{\Theta}_{(cell)} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2}$Here,$n = 2$ (number of electr

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$0.914$

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(A) Applying the Nernst equation:$E_{(cell)} = E^{\Theta}_{(cell)} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^{+}]^2}$Here,$n = 2$ (number of electrons transferred).$E_{(cell)} = 1.05 - \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}$$E_{(cell)} = 1.05 - 0.02955 \log \frac{0.16}{0.000004}$$E_{(cell)} = 1.05 - 0.02955 \log (4 	imes 10^4)$$E_{(cell)} = 1.05 - 0.02955 (4 + \log 4)$$E_{(cell)} = 1.05 - 0.02955 (4 + 0.6021)$$E_{(cell)} = 1.05 - 0.02955 (4.6021)$$E_{(cell)} = 1.05 - 0.136 = 0.914 \, V$

Calculate the $emf$ of the cell in which the following reaction takes place:
$Ni_{(s)} + 2Ag^{+}(0.002 \, M) \rightarrow Ni^{2+}(0.160 \, M) + 2Ag_{(s)}$
Given that $E^{\Theta}_{(cell)} = 1.05 \, V$ (in $, V$)

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Calculate the $emf$ of the cell in which the following reaction takes place:$Ni_{(s)} + 2Ag^{+}(0.002 \, M) \rightarrow Ni^{2+}(0.160 \, M) + 2Ag_{(s)}$Given that $E^{\Theta}_{(cell)} = 1.05 \, V$ (in $, V$)