The cell potential for the given galvanic cell at 298 K is 1.38 V . Zn(s) | Zn ²⁺(C₁) Ag ⁺(2 10⁻³ M ) | Ag(s) The concentration of Zn ²⁺ ions ( C₁ ) in the solution is _____ M . (Given: E^ _ Zn ²⁺/ Zn = -0.76 V , E^ _ Ag ⁺/ Ag = 0.80 V and 2.303RT F = 0.06 V )

Step-by-step solution

The standard cell potential is calculated as: E^ _ cell = E^ _ cathode - E^ _ anode E^ _ cell = 0.80 - (-0.76) = 1.56 V The overall cell reaction is: Zn(s) + 2 Ag ⁺( aq ) Zn ²⁺( aq ) + 2 Ag(s) Here, the number of electrons transferred, n = 2 . Using the Nernst equation: E_ cell = E^ _ cell - 2.303RT nF Q 1.38 = 1.56 - 0.06 2 ( [ Zn ²⁺] [ Ag ⁺]^2 ) 1.38 - 1.56 = -0.03 ( C₁ (2 10⁻³)^2 ) -0.18 = -0.03 ( C₁ 4 10⁻⁶ ) ( C₁ 4 10⁻⁶ ) = 6 Taking antilog on both sides: C₁ 4 10⁻⁶ = 10^6 C₁ = 4 10⁻⁶ 10^6 = 4 M Answer: 4