Consider the following reduction processes:

${Zn}^{2 +} + 2{e}^{-} \to Zn ( s ) ; {E}^{o} = - 0 .76 V$

${Ca}^{2 +} + 2{e}^{-} \to Ca ( s ) ; {E}^{o} = - 2 .87 V$

${Mg}^{2 +} + 2{e}^{-} \to Mg ( s ) ; {E}^{o} = - 2 .36 V$

${Ni}^{2 +} + 2{e}^{-} \to Ni ( s ) ; {E}^{o} = - 0 .25 V$

The reducing power of the metals increases in the order:

Question

Consider the following reduction processes:

{Zn}^{2 +} + 2{e}^{-} \to Zn ( s ) ; {E}^{o} = - 0 .76 V

{Ca}^{2 +} + 2{e}^{-} \to Ca ( s ) ; {E}^{o} = - 2 .87 V

{Mg}^{2 +} + 2{e}^{-} \to Mg ( s ) ; {E}^{o} = - 2 .36 V

{Ni}^{2 +} + 2{e}^{-} \to Ni ( s ) ; {E}^{o} = - 0 .25 V

The reducing power of the metals increases in the order:

Quantrex Academy · Accepted Answer

Correct answer: C. Ni < Zn < Mg < Ca

Lower the standard reduction potential more will be the reducing power i.e gets self oxidised. Therefore the correct order is: Ni < Zn < Mg < Ca

Consider the following reduction processes: Zn ^ 2 + + 2 e ^ - → Zn...