Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
\(\begin{array}{rlr}\text { (a) } & \mathrm{{Al}^{3+} / {Al}} & \mathrm{{E}_{RP}^{o}=-1.66} \\\text { (b) } & \mathrm{{Fe}^{2+} / {Fe}} & \mathrm{{E}_{RP}^{o}=-0.44} \\\text { (c) } & \mathrm{{Cu}^{2+} / {Cu}} & \mathrm{{E}_{RP}^{o}=+0.34} \\\text { (d) } & \mathrm{{F}_{2}({~g}) / 2 {~F}^{-}({aq})} & \mathrm{{E}_{RP}^{o}=+2.87}\end{array}\)

   
The correct choice among the above is:

1. (a, b)

2. (b, c)

3. (c, d)

4. (a, d)

The reducing ability of the metals K, Au, Zn, and Pb follows the order

${\mathrm{E}}^{\Theta }$ values of some redox couples are given below. On the basis of these values choose the correct option.

${\mathrm{E}}^{\mathrm{\Theta }}<mspace\; width="1em"></mspace>\mathrm{values}:<mspace\; width="1em"></mspace>{\mathrm{Br}}_2/{\mathrm{Br}}^{-}=+1.90$
${\mathrm{Ag}}^{+}/\mathrm{Ag}\left(\mathrm{s}\right)=+0.80$
${\mathrm{Cu}}^{2+}/\mathrm{Cu}\left(\mathrm{s}\right)=+0.34;<mspace\; width="1em"></mspace>$
${\mathrm{I}}_2\left(\mathrm{s}\right)/{\mathrm{I}}^{-}=+0.54$

The Mn³⁺ ion is unstable in solution and undergoes disproportionation reaction to give Mn²⁺, MnO₂ and H⁺ ion. The balanced ionic equation for the reaction is:

The correct arrangement of increasing order for reducing power of elements is:

The standard electrode potential (E°) values of  Al³⁺/ Al,  Ag⁺ / Ag, K⁺ / K,  and Cr³⁺ / Cr are –1.66 V, 0.80 V, –2.93 V, & –0.79 V respectively. The correct decreasing order of the reducing power of the metals is:

The correct statement about electrolysis of an aqueous solution of ${\mathrm{CuCl}}_2$ with Pt electrode is-