For the cell reaction 2Ce4- - Co - 2Ce3- - Co2-Eocell is 1-89 V- If EoCo2-Co is - 0-28 V- What is the value of Eo Ce4-Ce3-

We know, 2Ce^4+ + Co → 2Ce^3+ + Co^2+ E^o_cell = nbsp;E^o _Ce^4+/Ce^3++E^o_ Co/Co^2+ nbsp; ⇒ 1.89 = nbsp;E^o _Ce^4+/Ce^3+ + 0.28 ⇒ E^o _Ce^4+/Ce = 1.89 ...

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Byju's Answer

Standard VII

Chemistry

Equilibrium Constant from Nernst Equation

For the cell ...

Question

For the cell reaction $2 C e^{4 +} + C o \to 2 C e^{3 +} + C o^{2 +}$
$E_{c e l l}^{o} is 1.89 V$ . $If E_{C o^{2 +} | C o}^{o} is - 0.28 V$ , What is the value of $E_{C e^{4 +} | C e^{3 +}}^{o}$

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Solution

We know,
$2 C e^{4 +} + C o \to 2 C e^{3 +} + C o^{2 +}$
$E_{c e l l}^{o} = E_{C e^{4 +} / C e^{3 +}}^{o} + E_{C o / C o^{2 +}}^{o}$
$\Rightarrow 1.89 = E_{C e^{4 +} / C e^{3 +}}^{o} + 0.28$
$\Rightarrow E_{C e^{4 +} / C e}^{o} = 1.89 - 0.28 = 1.61 V$

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Similar questions

Q. 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}

Q. For the cell reaction,

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

;

E_{c e l l}^{o}

cell is

1.89 V

. If

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

- 0.28 V

, what is the value of

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

Q. Find the equilibrium constant for the reaction,

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

[Given:

E_{C e^{4 +} / C e^{3 +}}^{o} = 1.44 V; E_{F e^{3 +} / F e^{2 +}}^{o} = 0.68 V

]

Q. Calculate the equilibrium constant for the reaction.

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

, [given:

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

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

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 +}}^{⊖}

= 0.68 V.

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For the cell reaction 2Ce4++Co→2Ce3++Co2+ Eocell is 1.89 V. If EoCo2+|Co is−0.28 V, What is the value of EoCe4+|Ce3+

We know, 2Ce4++Co→2Ce3++Co2+ Eocell=EoCe4+/Ce3++EoCo/Co2+ ⇒1.89=EoCe4+/Ce3++0.28 ⇒EoCe4+/Ce=1.89−0.28=1.61 V

For the cell reaction $2 C e^{4 +} + C o \to 2 C e^{3 +} + C o^{2 +}$
$E_{c e l l}^{o} is 1.89 V$ . $If E_{C o^{2 +} | C o}^{o} is - 0.28 V$ , What is the value of $E_{C e^{4 +} | C e^{3 +}}^{o}$

We know,
$2 C e^{4 +} + C o \to 2 C e^{3 +} + C o^{2 +}$
$E_{c e l l}^{o} = E_{C e^{4 +} / C e^{3 +}}^{o} + E_{C o / C o^{2 +}}^{o}$
$\Rightarrow 1.89 = E_{C e^{4 +} / C e^{3 +}}^{o} + 0.28$
$\Rightarrow E_{C e^{4 +} / C e}^{o} = 1.89 - 0.28 = 1.61 V$