Question

Write the Nernst equation and emf of the following cells at 298 K:

(i) Mg(s)|Mg²⁺ (0.001 M)∥Cu²⁺ (0.0001 M)|Cu(s).
(ii) Fe(s)|Fe²⁺ (0.001 M)∥H⁺ (1M)|H₂ (g)(1bar)|Pt(s)
(iii) Sn(s)|Sn²⁺ (0.050 M)∥H⁺(0.020 M)|H₂ (g)(1bar)|Pt(s)
(iv) Pt(s)|Br₂ (l)|Br−(0.010 M)∥H⁺ (0.030 M)|H₂ (g)(1bar)|Pt(s).

Answer 1

The Nernst equation is used to calculate the EMF of electrochemical cells considering the standard cell potential, number of electrons involved, and the reaction quotient. At 298 K, the equation simplifies, making it easier to calculate the EMF for each given cell setup.

Step-by-step explanation:

To calculate the electromotive force (EMF) of the given electrochemical cells at 298 K, we can use the Nernst equation, which takes into account the standard electrode potential, temperature, number of electrons involved in the redox reaction, and the reaction quotient (Q). The Nernst equation in its general form is E = E° - (RT/nF)ln(Q), where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature in kelvin, n is the number of moles of electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient. At standard temperature (298 K), the equation simplifies to E = E° - (0.0592/n)log(Q).

Using the values provided in the question and knowing the temperature, we can substitute and solve for each cell to find the respective EMFs. The calculations would involve using the provided concentrations to find the reaction quotients for the cells and then applying the Nernst equation with the standard electrode potentials and the number of electrons transferred in each reaction.