Question
  1. Balance $\text{MnO}_4^-+\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{Mn}^{2+}$ in acidic medium by ion electron method. 
  2.  Given the standard electrode potentials:

$\frac{\text{K}_+}{\text{K}}=-2.93\text{V},$

$\frac{\text{Ag}^+}{\text{Ag}}=+0.80\text{V},$

$\frac{\text{Mg}^{2+}}{\text{Mg}}=-2.37\text{V}$

Arrange these metals in order of increasing reducing power.

Answer

  1. $\text{MnO}_4^-+\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{Mn}^{2+}$ in acidic medium

$[\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-]\times5\ \cdots\text{(i)}$

$5\text{e}^-+8\text{H}^++\text{MnO}_4^-\xrightarrow{ \ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O} \ \cdots\text{(ii)}$

Adding (i) and (ii), we get 

$5\text{Fe}^{2+}+8\text{H}^++\text{MnO}_4^-\xrightarrow{ \ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O} \ +5\text{Fe}^{3+}$

  1. Ag < Mg < K is increasing order of reducing power.

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