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Electrochemical series; Ni galvanic cell emf

Short answer

The electrochemical series ranks electrodes by increasing standard reduction potential. The Ni anode is paired with a Cu || Cu²⁺ (1 M) cathode. Cell EMF = 0.59 − (0.0591/2) log(1.5/1) ≈ 0.585 V.

What is an electrochemical series? With the help of electrochemical series complete the following galvanic cell and calculate the cell emf. (−) Ni / Ni²⁺(1.5M) // [?] E°Ni²⁺/Ni = −0.25 V

Electrochemical series

The arrangement of elements (electrodes) in increasing order of their standard reduction potentials (or decreasing order of oxidation potential) is the electrochemical series. Metals at the top (Li, K, Zn …) are strong reducing agents and lose electrons easily; metals at the bottom (Cu, Ag, Au) are weak reducing agents. A metal higher in the series displaces a lower metal from its salt solution.

Part of the series (E° reduction, V): Zn −0.76 < Ni −0.25 < H₂ 0.00 < Cu +0.34 < Ag +0.80.

Completing the cell

Ni (E° = −0.25 V) is the anode, so the right-hand electrode must have a higher reduction potential. Choosing copper (E° = +0.34 V), with 1 M Cu²⁺ (concentration not given, so standard assumed):

(−) Ni | Ni²⁺ (1.5 M) || Cu²⁺ (1 M) | Cu (+)

Anode   : Ni → Ni²⁺ + 2e⁻
Cathode : Cu²⁺ + 2e⁻ → Cu
Cell    : Ni + Cu²⁺ → Ni²⁺ + Cu         (n = 2)

E°cell = 0.34 − (−0.25) = 0.59 V
Q = [Ni²⁺]/[Cu²⁺] = 1.5/1 = 1.5 ,  log Q = 0.176
E = 0.59 − 0.02955 × 0.176 = 0.59 − 0.0052 = 0.585 V

Answer: EMF ≈ 0.585 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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