Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

Calculate the cell potential for $Ag_{(s)}|Ag^{+} \, (0.01 \ M)||Ag^{+} \, (0.1 \ M)|Ag_{(s)}$ at $298 \ K$. (in $V$)

For the following electrochemical cell at $298 \ K$,
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is

For the cell reaction $Zn_{(s)} + Ni^{2+}_{(aq)} (a = 0.1) \rightarrow Zn^{2+}_{(aq)} (a = 1.0) + Ni_{(s)}$ at $298 \ K$,the $emf$ is observed to be $0.5105 \ V$. What is the standard $emf$ $(E^{\circ}_{cell})$ of the cell in $V$?

Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$
(Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)