Question

For a galvanic cell that uses the following two half-reactions,Cr 2O 7 2-( aq) 14 H ( aq) 6 e - → 2 Cr 3 ( aq) 7 H 2O( l)Sn( s) → Sn 2 ( aq) 2 e -how many moles of Sn(s) are oxidized by four moles of Cr 2O 7 2-

Answer 1

Answer : The number of moles of Sn(s) oxidized are 12 moles.

Explanation :

The given half cell reactions are:

Reduction half reaction : $Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{3+}(aq)+7H_2O(l)$

Oxidation half reaction : $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-$

In order to balance the electrons, we will multiply oxidation half reaction by 3, we get:

Reduction half reaction : $Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{3+}(aq)+7H_2O(l)$

Oxidation half reaction : $3Sn(s)\rightarrow 3Sn^{2+}(aq)+6e^-$

Now adding both half reaction, we get the overall reaction.

$Cr_2O_7^{2-}(aq)+14H^+(aq)+3Sn(s)\rightarrow 2Cr^{3+}(aq)+3Sn^{2+}(aq)+7H_2O(l)$

From the overall complete reaction, we conclude that

As, 1 mole of $Cr_2O_7^{2-}$ will oxidizes 3 moles of Sn

So, 4 moles of $Cr_2O_7^{2-}$ will oxidizes $4\times 3=12$ moles of Sn

Therefore, the number of moles of Sn(s) oxidized are 12 moles.