The standard reduction potentials at $$298 \mathrm{~K}$$ for the following half cells are given below :

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \quad \mathrm{E}^{\circ}=1.33 \mathrm{~V}$$

$$\begin{array}{ll} \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} & \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} & \mathrm{E}^{\circ}=1.40 \mathrm{~V} \end{array}$$

Consider the given electrochemical reactions,

The number of metal(s) which will be oxidized be $$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$$, in aqueous solution is _________.

The amount of electricity in Coulomb required for the oxidation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ is __________ $\times 10^5 \mathrm{C}$.

Consider the following redox reaction :

$$ \mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $$

The standard reduction potentials are given as below $\left(\mathrm{E}_{\text {red }}^0\right)$ :

$$ \begin{aligned} & \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\\\ & \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V} \end{aligned} $$

If the equilibrium constant of the above reaction is given as $\mathrm{K}_{\mathrm{eq}}=10^x$, then the value of $x=$ __________ (nearest integer)

The potential for the given half cell at $298 \mathrm{~K}$ is (-) __________ $\times 10^{-2} \mathrm{~V}$

$$ \begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\\\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}} \end{aligned} $$

(Given : $2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )