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* 3=12` moles of Sn

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