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Sagot :
half-reactions
cathode(reduction) : Ni2+(aq) + 2e → Ni(s)
anode(oxidation) : Mg (s) → Mg²⁺ (aq) + 2e−
Further explanation
Given
Ni and Mg reaction
Required
the half-reactions
Solution
we determine which is the more reduced / oxidized of the two elements by looking at the voltaic series or the standard potential value
In voltaic series
Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
If we look at the Mg metal located to the left of the Ni metal, so the Mg metal is more easily oxidized and can reduce the Ni metal which is on the right
Mg oxidation
Mg(s) – Mg2+(aq) + 2e
Ni reduction
Ni2+(aq) + 2e → Ni(s)
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