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Question: How many moles of \(K_{2}Cr_{2}O_{7}\) can be reduced by 1 mole of\(Sn^{2 +}\)....

How many moles of K2Cr2O7K_{2}Cr_{2}O_{7} can be reduced by 1 mole ofSn2+Sn^{2 +}.

A

1/3

B

1/6

C

2/3

D

1

Answer

1/3

Explanation

Solution

Cr2O72−+14H++6e−→2Cr3++7H2OCr_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3 +} + 7H_{2}O

(Sn2+→Sn4++2e−)×3(Sn^{2 +} \rightarrow Sn^{4 +} + 2e^{-}) \times 3

Cr2O72−+14H++3Sn2+→3Sn4++2Cr3++7H2O‾\overline{Cr_{2}O_{7}^{2 -} + 14H^{+} + 3Sn^{2 +} \rightarrow 3Sn^{4 +} + 2Cr^{3 +} + 7H_{2}O}

It is clear from this equation that 3 moles of Sn2+Sn^{2 +} reduce one mole of Cr2O72−Cr_{2}O_{7}^{2 -}, hence 1 mol. of Sn2+Sn^{2 +} will reduce 13\frac{1}{3} moles of Cr2O72−Cr_{2}O_{7}^{2 -}.