Find the observed $EMF$ of the cell $Cd | Cd^{2+}(0.01 \ M) || Cu^{2+}(0.01 \ M) | Cu$ under conditions where the internal resistance is $4 \ \Omega$ and it is producing a current of $0.15 \ A$. (Given: $E^{\circ}_{Cu^{2+}/Cu} = 0.35 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.4 \ V$) (in $V$)

Which of the following is the correct expression for the electrode potential of a cell?

The standard electrode potential of a $Cu^{2+} | Cu$ electrode is $0.34 \, V$ (reduction potential). What will be the electrode potential of a $0.001 \, M \, Cu^{2+}$ solution in $V$?

For a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

In the given electrochemical cell, $Ag_{(s)} | AgCl_{(s)} | Cl^-_{(aq)}, Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} | Pt_{(s)}$ at $298 \ K$, the cell potential $(E_{cell})$ will increase when :
$(A)$ Concentration of $Fe^{2+}$ is increased.
$(B)$ Concentration of $Fe^{3+}$ is decreased.
$(C)$ Concentration of $Fe^{2+}$ is decreased.
$(D)$ Concentration of $Fe^{3+}$ is increased.
$(E)$ Concentration of $Cl^-$ is increased.
Choose the correct answer from the options given below :

Which of the following relations represents the correct relation between standard electrode potential and equilibrium constant?
$I$. $\log K = \frac{nF E^o}{2.303 RT}$
$II$. $K = e^{\frac{nF E^o}{RT}}$
$III$. $\log K = -\frac{nF E^o}{2.303 RT}$
$IV$. $\log K = 0.4342 \frac{nF E^o}{RT}$
Choose the correct statement$(s)$.