Question

For the following electron-transfer reaction:

2Ag^+(aq) + Mn(s) rightarrow 2Ag(s) + Mn^2+(aq)
1. The oxidation half-reaction is:_______
2. The reduction half-reaction is:_______

Answer 1

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)

Step-by-step explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻