Simulation · Chemistry · Class 12

How does concentration change a cell's voltage?

From the lesson The Nernst equation in Electrochemistry. Change the values and watch what happens.

The idea behind it

NCERT §2.3

For Mⁿ⁺(aq) + ne⁻ → M(s), the potential at a general concentration is E = E° − (RT/nF) ln(1/[Mⁿ⁺]), since a pure solid is taken as unity. R = 8.314 J K⁻¹ mol⁻¹, F = 96487 C mol⁻¹, T in kelvin.
For the Daniell cell, E_cell = E°_cell − (RT/2F) ln([Zn²⁺]/[Cu²⁺]). The cell voltage rises when [Cu²⁺] is raised or [Zn²⁺] lowered.
At 298 K, with base-10 logs, RT/F × 2.303 = 0.059 V, so E_cell = E°_cell − (0.059/n) log Q.
For aA + bB + ne⁻ → cC + dD, Q = [C]^c[D]^d/[A]^a[B]^b, with pure solids and liquids left out. Use the same n for both electrodes: for Ni(s) + 2Ag⁺ → Ni²⁺ + 2Ag(s), n = 2 and Q = [Ni²⁺]/[Ag⁺]².
Worked: Mg|Mg²⁺(0.130 M)||Ag⁺(0.0001 M)|Ag with E° = 3.17 V gives E = 3.17 − (0.059/2) log(0.130/10⁻⁸) = 3.17 − 0.21 = 2.96 V.
A hydrogen electrode in a solution of pH 10 (at 1 bar H₂): E = −0.059 × pH = −0.59 V.

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