You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Nernst Equation

If the cell v...

Question

If the cell voltage is 1.23 V for the $H_{2} - O_{2}$ fuel cell and for the
half cell :
$O_{2} + 2 H_{2} O + 4 e^{-} ⇌ 4 ⊖ O H$ has $E^{⊖} = 0.40 V$ , then $E^{⊖}$ for $2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 ⊖ O H$ will be:

0.56 V

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

0.73 V

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

0.41 V

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

0.83 V

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

Open in App

Solution

The correct option is C $0.83 V$
$\Rightarrow E_{c e l l}^{⊖} = 1.23 = 0.4 - (E_{r e d}^{⊖})_{a} \Rightarrow (E_{r e d}^{⊖})_{a} = - 0.83 V$

Suggest Corrections

Similar questions

Q. In

H_{2} - O_{2}

fuel cell the reaction occuring at cathode is:

In H₂ - O₂ fuel cell the reaction occuring at cathode is :

Q. In a fuel cell

H_{2}

and

O_{2}

react to product electricity. In the process, hydrogen is oxidized at anode and oxygen at cathode. If 67.2 l of

H_{2}

at STP reacts in 15 min, the average current produced (

i n A

) is_______. (nearest value)

Anode :

H_{2} + 2 ⊖ O H \to 2 H_{2} O + 2 e^{-}

Cathode :

O_{2} + 2 H_{2}) + 2 e^{-} \to 4 ⊖ O H

In a fuel cell, hydrogen and oxygen react to produce electricity. In the process, hydrogen gas is oxidized at anode and oxygen at cathode. If 67.2 L of $H_{2}$ at STP reacts in 15 min, the average current (in A) produced is :

Anode reaction : $H_{2} + 2 ⊖ O H \to 2 H_{2} O + 2 e^{-}$

Cathode reaction :

O_{2} + 2 H_{2} O + 2 e^{-} \to 4 ⊖ O H

(write your answer to nearest integer)

Q. Assertion :

(H_{2} + O_{2})

fuel cell gives a constant voltage throughtout its life. Reason: In this fuel cell,

H_{2}

reacts with

⊖ O H

ions, yet the over all

[⊖ O H]

does not change.

Join BYJU'S Learning Program

If the cell voltage is 1.23 V for the H2−O2 fuel cell and for thehalf cell :O2+2H2O+4e−⇌4⊖OH has E⊖=0.40V, then E⊖ for 2H2O+2e−⇌H2+2⊖OH will be:

The correct option is C 0.83V⇒E⊖cell=1.23=0.4−(E⊖red)a⇒(E⊖red)a=−0.83V

If the cell voltage is 1.23 V for the $H_{2} - O_{2}$ fuel cell and for the
half cell :
$O_{2} + 2 H_{2} O + 4 e^{-} ⇌ 4 ⊖ O H$ has $E^{⊖} = 0.40 V$ , then $E^{⊖}$ for $2 H_{2} O + 2 e^{-} ⇌ H_{2} + 2 ⊖ O H$ will be:

The correct option is C $0.83 V$
$\Rightarrow E_{c e l l}^{⊖} = 1.23 = 0.4 - (E_{r e d}^{⊖})_{a} \Rightarrow (E_{r e d}^{⊖})_{a} = - 0.83 V$