Given : $\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V}$

$\mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V}$

$\begin{aligned} & \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned}$

When the potential of the cell is $0.712 \mathrm{~V}$ at $298 \mathrm{~K}$, the ratio $\left[\mathrm{Fe}^{2+}\right] /\left[\mathrm{Fe}^{3+}\right]$ is _____________. (Nearest integer)

Given : $\mathrm{Fe}^{3+}+\mathrm{e}^{-}=\mathrm{Fe}^{2+}, \mathrm{E}^{\theta} \mathrm{Fe}^{3+}, \mathrm{Fe}^{2+} \mid \mathrm{Pt}=0.771$

$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$

Given : $\mathrm{\frac{2.303RT}{F}=0.059~V}$

The $\mathrm{E_{cell}}$ for the given cell is 0.1115 V at 298 K when ${{\left[ {{M^ + }(aq)} \right]} \over {\left[ {{M^{3 + }}(aq)} \right]}} = {10^a}$

The value of $a$ is ____________

Given : $\mathrm{E_{{M^{3 + }}/{M^ + }}^\theta = 0.2}$ V

${{2.303RT} \over F} = 0.059V$

Given : $\mathrm{C{r_2}O_7^{2 - } \to C{r^{3 + }}\,;\,E^\circ = 1.330}$V

and $\mathrm{{{2.303\,RT} \over F} = 0.059}$ V

The potential for the half cell reaction is $x\times10^{-3}$ V. The value of $x$ is __________

Given : $\mathrm{K}_{\mathrm{sp}} \mathrm{Cu}(\mathrm{OH})_{2}=1 \times 10^{-20}$

Take $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$

The reduction potential at $\mathrm{pH}=14$ for the above couple is $(-) x \times 10^{-2} \mathrm{~V}$. The value of $x$ is ___________

[Given : $E_{C{u^{2 + }}/Cu}^\Theta = 0.34\,V$ ; $E_{S{n^{2 + }}/Sn}^\Theta = - 0.14\,V$ ; F = 96500 C mol^$-$1]

(Given atomic number $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$)

[Given : $\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13}$ at $298 \mathrm{~K}$

$\begin{aligned} & \lambda_{\mathrm{Ag}^{+}}^{0}=6 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{Br}^{-}}^{0}=8 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \left.\lambda_{\mathrm{NO}_{3}^{-}}^{0}=7 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\right] \end{aligned}$