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Ionization Energies

For the cell ...

Question

For the cell reaction $2 {C e}^{4 +} + C o ⟶ 2 {C e}^{3 +} + {C o}^{2 +}$
$E_{c e l l}^{0}$ is 1.89 V. If $E_{{C o}^{2 +} | C o}^{0}$ is -0,28 V, what is the value of $E_{{C o}^{4 +} | {C e}^{3 +}}^{0}$ ?

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Solution

$2 {C e}^{4 +} + C o \to 2 {C e}^{3 +} + {C o}^{2 +}$
Anode : $C o \to {C o}^{2 +} + {2 e}^{-}$
Cathode : ${2 C e}^{4 +} + 2 e^{-} \to {C e}^{3 +}$
$E_{c e l l} = E_{R} - E_{L}$
$E_{c e l l} = E_{{C e}^{4 +} / {C e}^{3 +}} - E_{C o^{2 +} / C o}$
$1.89 = E_{{C e}^{4 +} / {C e}^{3 +}} - (- 0.28)$
$E_{{C e}^{4 +} / {C e}^{3 +}} = 1.89 - 0.28$
$E_{{C e}^{4 +} / {C e}^{3 +}} = 1.61 V$

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Similar questions

2 C e^{4 +} + C o \to 2 C e^{3 +} + C o^{2 +}

E_{c e l l}^{o} is 1.89 V

If E_{C o^{2 +} | C o}^{o} is - 0.28 V

E_{C e^{4 +} | C e^{3 +}}^{o}

2 {C e}^{4 +} + C o \to 2 {C e}^{3 +} + {C o}^{3 +}

E_{c e l l}^{o}

1.89 V

E_{{C o}^{3 +} / C o}

- 0.28 V

E_{{C e}^{4 +} / {C e}^{3 +}}^{o}

F e^{2 +} + C e^{4 +} | F e^{3 +} + C e^{3 +}

E^{0} C e^{4 +} / C e^{3 +} = 1.44

E^{0} F e^{3 +} / F e^{2 +} = 0.68 V]

E_{c e l l}^{0}

- 0.32 V

25^{o} C

F e (s) + Z n^{2 +} (a q) ⇌ Z n (s) + F e^{2 +} (a q)

log [F e^{2 +}]

1 M

Z n^{2 +}

The equilibrium constant for the reaction is:

$2 F e^{3 +} + 3 I^{⊖} ⇌ F e^{3 +} + C e^{3 +}$

Given :

$E_{C e^{4 +} 1 C e^{3 +}}^{⊖}$ = 1.44 V

E_{F e^{3 +} 1 F e^{2 +}}^{⊖}

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