Answer:
Option D.
2Ag⁺(aq) + Ni(s) —> 2Ag(s) + Ni²⁺(aq)
Explanation:
From the question given above:
Ni(s) —> Ni²⁺(aq) + 2e¯
Ag⁺(aq) + e¯ —> Ag(s)
We can obtain the net ionic equation by combining both equation as shown below:
Ag⁺(aq) + e¯ + Ni(s) —> Ag(s) + Ni²⁺(aq) + 2e¯
Next, we shall balance the equation. This can be obtained as follow:
Ag⁺(aq) + e¯ + Ni(s) —> Ag(s) + Ni²⁺(aq) + 2e¯
There are 2 atoms of e¯ on the right side and 1 atom on the left side. It can be balance by putting 2 in front of e¯ as shown below:
Ag⁺(aq) + 2e¯ + Ni(s) —> Ag(s) + Ni²⁺(aq) + 2e¯
Next, we shall balance the charge on both sides. This is illustrated below:
TOTAL charge on the left = +1 + (–2)
= 1 – 2
= –1
TOTAL charge on the right side = +2 + (–2)
= 2 – 2
= 0
Thus, to balance the charge, put 2 in front of Ag⁺ as shown below:
2Ag⁺(aq) + 2e¯ + Ni(s) —> Ag(s) + Ni²⁺(aq) + 2e¯
There are 2 atoms of Ag on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Ag as shown below:
2Ag⁺(aq) + 2e¯ + Ni(s) —> 2Ag(s) + Ni²⁺(aq) + 2e¯
Next, cancel out 2e¯ from both side to obtain the net ionic equation. Thus, we have:
2Ag⁺(aq) + Ni(s) —> 2Ag(s) + Ni²⁺(aq)
