The standard reaction- Gibbs energy for the electrochemical reaction- Cr 2 O 72-6 Fe 2-14 H - 2 Cr 3-6 F e 3-7 H 2 O is -793 kJ mol -1 The standard cell potential is-A- -1-37 VB- -4-11 VC -2-74 VD- -2-05 V

The correct option is A + 1.37 VΔ G^∘ = nFE^∘_cell Here nbsp; n = 6 793 × 10^3 = 6 × 96500 × E^∘_cell E^∘_cell =+1.3696 = 1.37V ...

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Standard XII

Chemistry

Enthalpy of Solution

The standard ...

Question

The standard reaction, Gibbs energy for the electrochemical reaction.
$C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O i s - 793 k J m o l^{- 1}$ The standard cell potential is:

+ 1.37 V

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+4.11 V

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+2.74 V

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+2.05 V

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Solution

The correct option is A + 1.37 V
$Δ G^{\circ} = - n F E_{c e l l}^{\circ}$
Here n = 6
$- 793 \times 10^{3} = - 6 \times 96500 \times E_{c e l l}^{\circ} E_{c e l l}^{\circ} = + 1.3696$
= 1.37V

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Similar questions

The standard reaction gibbs energy for the electrochemical reaction.

$C r_{2} O_{7}^{2 -} + 2 F e + 14 H^{+} \to 2 C r^{3 +} + 2 F e^{3 +} + 2 H_{2} O$ is $- 793 k J m o l^{- 1}$ . The standard cell potential is:

The standard reaction Gibbs energy for the electrochemical reaction,

$C r_{2} O_{7}^{2 -} + 2 F e + 14 H^{+} \to 2 C r^{3 +} + 2 F e^{3 +} + 2 H_{2} O$ , is $- 793 k J m o l^{- 1}$ . The standard cell potential is Volts.

Write half-reactions using electrons.

{C r}_{2} O_{2}^{7 -} + 6 {F e}^{2 +} + 14 H^{+} ⟶ 2 {C r}^{3 +} + 6 {F e}^{3 +} + 7 H_{2} O

C r_{2} O_{7}^{2 -} (a q, 1 M) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q, 1 M) + 7 H_{2} O

. The standard electrode potential for the following reaction is + 1.33V. what is the potential at

p H = 2.0

Q. Given a reaction:

{C r}_{2} O_{7}^{2 -} + {F e}^{2 +} + H^{\oplus} ⟶ {C r}^{3 +} + {F e}^{3 +} + H_{2} O

The balanced equation is

6 {F e}^{2 +} + 14 H^{\oplus} + {C r}_{2} O_{7}^{2 -} ⟶ 6 {F e}^{3 +} + 2 {C r}^{3 +} + 7 H_{2} O

.
If true enter 1, else enter 0.

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The standard reaction, Gibbs energy for the electrochemical reaction. Cr2O2−7+6Fe2++14H+→2Cr3++6Fe3++7H2O is −793 kJmol−1 The standard cell potential is:

The correct option is A + 1.37 VΔG∘=−nFE∘cell Here n = 6 −793×103=−6×96500×E∘cellE∘cell=+1.3696 = 1.37V

The standard reaction, Gibbs energy for the electrochemical reaction.
$C r_{2} O_{7}^{2 -} + 6 F e^{2 +} + 14 H^{+} \to 2 C r^{3 +} + 6 F e^{3 +} + 7 H_{2} O i s - 793 k J m o l^{- 1}$ The standard cell potential is:

The correct option is A + 1.37 V
$Δ G^{\circ} = - n F E_{c e l l}^{\circ}$
Here n = 6
$- 793 \times 10^{3} = - 6 \times 96500 \times E_{c e l l}^{\circ} E_{c e l l}^{\circ} = + 1.3696$
= 1.37V