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

The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

What will be the electromotive force of the following cell in $V$?
$Fe|Fe^{+2}(0.2 \ M)||Au^{+3}(0.02 \ M)|Au$
Given: $E_{Fe^{+2}/Fe}^o = -0.44 \ V$ and $E_{Au^{+3}/Au}^o = 1.50 \ V$

Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?