Question

A voltaic cell utilizes the following reaction and operates at 298 K. 3 Ce4+(aq) + Cr(s) 3 Ce3+(aq) + Cr3+(aq)

(a) What is the emf of this cell under standard conditions? V

(b) What is the emf of this cell when [Ce4+] = 2.0 M, [Ce3+] = 0.014 M, and [Cr3+] = 0.011 M? V

(c) What is the emf of the cell when [Ce4+] = 0.56 M, [Ce3+] = 0.82 M, and [Cr3+] = 1.3 M? V

Answer 1

A voltaic cell utilizes the following reaction,

3Ce⁴⁺ (aq.) + Cr(s) ----------> 3Ce³⁺(aq.) + Cr³⁺(aq.)

Oxidation Half reaction: Cr (s) -------> Cr³⁺ (aq.) + 3e^-    E⁰red (Cr3+/Cr) = -0.74V

Reduction Half reaction: 3Ce⁴⁺ (aq.) + 3e^- --------> 3Ce³⁺ (aq.) E⁰red (Ce4+/Ce3+) = +1.78V

T = 298 K.

Number of electrons transferred (n) = 3

(a) EMF of this cell under standard conditions

Under standard conditions i.e. at unit concentration (activity more precisely) of all ions involved EMF of cell is called as Standard EMF of the cell,

Ecell = E⁰cell = |E⁰red (Ce⁴⁺/Ce³⁺) - E⁰red (Cr³⁺/Cr)|

E⁰cell =|(+1.78V) – (-0.74V)|

E⁰cell = +2.52 V

Nernst equation for non-standard condition and at T=298 K

Ecell = E⁰cell – (0.0591/n) log{[ Ce³⁺]³[ Cr³⁺]/[ Ce⁴⁺]³}

With n = 3 and E⁰cell = +2.52 V

Ecell = (+2.52) – (0.0591/3) log{[ Ce³⁺]³[ Cr³⁺]/[ Ce⁴⁺]³}

Ecell = (+2.52) – (0.0197) log{[ Ce³⁺]³[ Cr³⁺]/[ Ce⁴⁺]³} …….. (1)

Using this equation,

(b) EMF of this cell when [Ce⁴⁺] = 2.0 M, [Ce³⁺] = 0.014 M, and [Cr³⁺] = 0.011 M

Put these values in eq. (1)

Ecell = (+2.52) – (0.0197) log{(0.014)³(0.011) /2³}

Ecell = (+2.52) – (0.0197) x (-8.423)

Ecell = (+2.52) –(0.166)

Ecell = +2.686 V

(c)EMF of the cell when [Ce⁴⁺] = 0.56 M, [Ce³⁺] = 0.82 M, and [Cr³⁺] = 1.3 M

Put these values in eq. (1)

Ecell = (+2.52) – (0.0197) log{(0.82)³(1.3) /(0.56)³}

Ecell = (+2.52) – (0.0197) x (0.612)

Ecell = (+2.52) –(0.0121)

Ecell = +2.508 V

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