The standard $emf$ of a galvanic cell involving $3$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the reaction of the cell is:

Calculate the $emf$ of the cell at $25^{\circ} C$.
Cell notation: $M | M^{2+} (0.01 \ M) || M^{2+} (0.0001 \ M) | M$
Given: $E_{cell}^{o} = 4 \ V$ and $\frac{RT}{F} \ln 10 = 0.06$. (in $V$)

In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

The value of the reaction quotient $(Q)$ for the cell $Zn_{(s)} | Zn^{2+}(0.01 \ M) || Cu^{2+}(1.25 \ M) | Cu_{(s)}$ is:

If the standard electrode potential of $Cu^{2+}/Cu$ electrode is $0.34 \, V$,what is the electrode potential of $0.01 \, M$ concentration of $Cu^{2+}$ $(T = 298 \, K)$ (in $, V$)?

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$