What are the half-reactions for a galvanic cell with Ni and Mg electrodes? A. Ni(s) → Ni2+(aq) + 2e and Mg2+(aq) + 2e → Mg(s)
B. Ni2+(aq) + 2e → Ni(s) and Mg(s) – Mg2+(aq) + 2e
C. Ni(s) Ni2+(aq) + 2e and Mg(s) → Mg2+(aq) + 2e
D. Ni2+(aq) + 2e → Ni(s) and Mg2+ (aq) + 2e → Mg(s)

half-reactions

cathode(reduction) : Ni2+(aq) + 2e → Ni(s)

anode(oxidation) : Mg (s) → Mg²⁺ (aq) + 2e−

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)

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm