Answer: the slope of the calibration curve is - 0.0295
Explanation:
as things are not under standard conditions that is; 1M of each reactants at T = 25°C and P = 1 atm; therefore we will use the Nernst Equation.
E = a × pNi + b
a is the slope
pNi = -log [Ni²⁺] = log 1/[Ni²⁺]
therefore the Half reaction;
Ni²⁺ (aq) + 2e⁻ = Ni (s)
here; E = E₀ + RT/zF ln[Ni²⁺]
we know that;
gas constant R = (8.31 J/mol-K)
absolute temperature T =25°C = 298 K
number of moles of electrons transferred by the cell's reaction z = 2
F is Faraday constant, = 9.64853399(24)×10⁴ C mol−1
E₀ = b
RT/zF ln[Ni²⁺] = alog 1/[Ni²⁺]
Logx = Ln x / Ln 10 = Ln (x/2.3)
so
a = (RT/zF ln[Ni²⁺])/log 1/[Ni²⁺]
= (RT/zF ln[Ni²⁺]) / ((ln 1/[Ni²⁺]) / 2.3)
= (2.3RT/zF ln[Ni²⁺]) / (ln 1/[Ni²⁺])
= (2.3RT/zF ln[Ni²⁺]) / (-ln [Ni²⁺])
= -2.3RT/zF
so we substitute
= (-2.3 × 8.31 ×298 ) / (2 × 9.64853399(24) × 10⁴)
= - 0.0295
therefore the slope of the calibration curve is - 0.0295
