How many moles of NaCl are present in a solution with a molarity of 8.59 M and a volume of 125 mL
2 answers:
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Answer:
(a) Cu²⁺ +2e⁻ ⇌ Cu
(c) 0.07 V
Explanation:
(a) Cu half-reaction
Cu²⁺ + 2e⁻ ⇌ Cu
(c) Cell voltage
The standard reduction potentials for the half-reactions are+
<u> E°/V </u>
Cu²⁺ + 2e⁻ ⇌ Cu; 0.34
Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241
The equation for the cell reaction is
E°/V
Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34
<u>2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; </u> <u>-0.241 </u>
Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10
The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation
(ii) Calculations:
T = 25 + 273.15 = 298.15 K