Question

Balance the following redox equation in acidic solution using the smallest integers possible and select the correct coefficient for the H+(aq) ion.

Cr2O72–(aq) + Sn2+(aq) → Cr3+(aq) + Sn4+(aq)

(A) 1 (no coefficient written)
(B) 2
(C) 3
(D) 4
(E) More than 4

Answer 1

The balanced redox reaction requires a coefficient of 14 for the H+ ion, making the correct answer (E) More than 4.

Step-by-step explanation:

To balance the redox equation Cr2O72−(aq) + Sn2+(aq) → Cr3+(aq) + Sn4+(aq) in an acidic solution, one must first separate the equation into two half-reactions and then balance each half-reaction for mass and charge.

The oxidation half-reaction is:
Sn2+ → Sn4+ + 2e−

The reduction half-reaction is:
Cr2O72− + 14H+ + 6e− → 2Cr3+ +7H2O

To balance the electrons between the oxidation and reduction half-reactions, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1. Finally, we combine the balanced half-reactions and add the coefficients to get the overall balanced equation.

The balanced redox reaction in acidic solution is:
Cr2O72−(aq) + 14H+(aq) + 3Sn2+(aq) → 2Cr3+(aq) + 3Sn4+(aq) + 7H2O(l)

Therefore, the correct coefficient for the H+(aq) ion is 14.

Answer 2

Coefficient of
`H^(+)(aq)` is more than 4

Step-by-step explanation:

Oxidation:
`Sn^(2+)(aq)\rightarrow Sn^(4+)(aq)`

• Balance charge:
  `Sn^(2+)(aq)-2e^(-)\rightarrow Sn^(4+)(aq)`......(1)

Reduction:
`Cr_(2)O_(7)^(2-)(aq)\rightarrow Cr^(3+)(aq)`

• Balance Cr:
  `Cr_(2)O_(7)^(2-)(aq)\rightarrow 2Cr^(3+)(aq)`
• Balance O and H in acidic medium:
  `Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)\rightarrow 2Cr^(3+)(aq)+7H_(2)O(l)`
• Balance charge:
  `Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-)\rightarrow 2Cr^(3+)(aq)+7H_(2)O(l)`.......(2)

`[3* Equation-(1)]+Equation(2)` gives balanced equation:

`3Sn^(2+)(aq)+Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)\rightarrow 3Sn^(4+)(aq)+2Cr^(3+)(aq)+7H_(2)O(l)`

So coefficient of
`H^(+)(aq)` is more than 4