For the cell reaction- Fes -Fe2- 0-1 M-H-1 M-H2 1 atm-PtsE0Fe-Fe2- - 0-44 V- the cell emf at 298 K is-

Fe(s) + 2H^+ (aq) → Fe^2+ (aq)+H_2(g) Equation is: E = E^0 0.0592 log[Fe^2+][P_H_2][H^+]^2 E^0 = E^0_(H^+/H_2 ) E_(Fe^2+/Fe )^0 = 0 ( 0.44) = 0.44 V ∴ E = ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Gibb's Free Energy

For the cell ...

Question

For the cell reaction: $F e (s) | F e^{2 +} (0.1 M) | | H^{+} (1 M) | H_{2} (1 atm) | P t (s)$
$E_{(F e / F e^{2 +})}^{0} = 0.44 V,$ the cell emf at $298 K$ is:

0.49 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.47 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.26 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.20 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.47 V$
$F e (s) + 2 H^{+} (a q) \to F e^{2 +} (a q) + H_{2} (g)$
Equation is: $E = E^{0} - \frac{0.059}{2} log \frac{[F e^{2 +}] [P_{H_{2}}]}{[H^{+}]^{2}} E^{0} = E_{(H^{+} / H_{2})}^{0} - E_{(F e^{2 +} / F e)}^{0} = 0 - (- 0.44) = 0.44 V ∴ E = 0.44 - \frac{0.059}{2} log \frac{0.1 \times 1}{1^{2}} ∴ E = 0.44 + 0.0295 = 0.4695 V$

Suggest Corrections

Similar questions

Q. Calculate the emf of the following cell at

298 K

F e (s) | F e^{2 +} (0.001 M) | | H^{+} (1 M) | H_{2} (g) (1 b a r), P t (s)

(Given:

E_{c e l l}^{\circ} = + 0.44 V

)

Q. The emf of the cell,

Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e

298 K

0.2905 V

then the value of equilibrium constant for the cell reaction is:

The EMF of the cell

$Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e$ at $298 K$ is $0.2905$ then the value of equilibrium constant for the cell reaction is :

Q. Will

F e_{(s)}

be oxidised to

F e^{2 +}

by the reaction with

1 M H C l

? (

E^{\circ}

for

F e / F e^{2 +} = + 0.44 V

)

Q. Calculate emf of the following cell at

25^{\circ} C

F e | F e^{2 +} (0.001 M) | H^{+} (0.01 M) | H_{2} (g) (1 b a r) P t (s) E^{\circ} (F e^{2 +} | F e) = - 0.44 V F^{\circ} (H^{+} / H_{2}) = 0.00 V

Join BYJU'S Learning Program

Explore more

Gibb's Free Energy

Standard XII Chemistry

Join BYJU'S Learning Program

For the cell reaction: Fe(s)|Fe2+(0.1 M)||H+(1 M)|H2(1 atm)|Pt(s) E0(Fe/Fe2+)=0.44 V, the cell emf at 298 K is:

The correct option is B 0.47 VFe(s)+2H+(aq)→Fe2+(aq)+H2(g) Equation is: E=E0−0.0592 log[Fe2+][PH2][H+]2E0=E0(H+/H2)−E0(Fe2+/Fe)=0−(−0.44)=0.44 V∴E=0.44−0.0592 log0.1×112∴E=0.44+0.0295=0.4695 V

For the cell reaction: $F e (s) | F e^{2 +} (0.1 M) | | H^{+} (1 M) | H_{2} (1 atm) | P t (s)$
$E_{(F e / F e^{2 +})}^{0} = 0.44 V,$ the cell emf at $298 K$ is: