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

Byju's Answer

Standard XII

Chemistry

EMF

If E⊖CH3COO...

Question

If $E_{C H_{3} C O O O H | C_{2} H_{5} O H}^{⊖} = 0.06 V$ , Then $E_{c e l l}^{⊖}$ of the reaction taking place in alcohol meter is :

1.39 V

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

1. 27 V

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

1.60 V

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

1.51 V

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

Open in App

Solution

The correct option is B $1. 27 V$
$E_{c e l l}^{⊖} = (E_{C r_{2}}^{⊖} O_{2}^{2 -} | C r^{3 +})_{r e d u c t i o n a t c a t h o d e}$
$- (E_{C H_{3}}^{⊖} C O O H | C_{2} H_{5} O H)_{r e d u c t i o n a t a n o d e}$
= 1.33 - 0.06 = 1.27 V

Suggest Corrections

Similar questions

Q. Given below are the half-cell reactions:

M n^{2 +} + 2 e^{-} \to M n; E^{\circ} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}); E^{\circ} = + 1.51 V

The

E^{\circ}

for

3 M n^{2 +} \to M n + 2 M n^{3 +}

will be:

Q. For the redox reaction,

Z n + C u_{(0.1 M)}^{2 +} \to Z n_{(1 M)}^{2 +} + C u

taking place in a cell

E_{c e l l}^{⊖} = 1.10

E_{C e l l}^{⊖}

for the galvanic cell will be?

Given below are the half-cell reaction:
$M n^{2 +} + 2 e^{⊖} \to M n; E^{⊖} = - 1.18 V$
$2 (M n^{3 +} + e^{⊖} \to M n^{2 +}; E^{⊖} = + 1.51 V$

The $E^{⊖}$ for $3 M n^{2 +} \to M n + 2 M n^{3 +}$ will be:
(IIT-JEE 2014)

Q. For the reaction,

Z n (s) + {C u}^{2 +} (0.1 M) ⟶ {Z n}^{2 +} (1 M) + C u (s)

taking place in a cell;

E_{c e l l}^{o}

1.10 V

E_{c e l l}

for the cell will be

(2.303 \frac{R T}{F} = 0.0591)

Q. Given below are the half-cell reactions

M n^{2 +} + 2 e^{-} \to M n, E^{o} = - 1.18 V

2 (M n^{3 +} + e^{-} \to M n^{2 +}), E^{o} = + 1.51 V

The

E^{o}

for

3 M n^{2 +} \to M n + 2 M n^{3 +}

will be:

Join BYJU'S Learning Program

If E⊖CH3COOOH|C2H5OH=0.06V, Then E⊖cell of the reaction taking place in alcohol meter is :

The correct option is B 1.27VE⊖cell=(E⊖Cr2O2−2|Cr3+)reductionatcathode−(E⊖CH3COOH|C2H5OH)reductionatanode = 1.33 - 0.06 = 1.27 V

If $E_{C H_{3} C O O O H | C_{2} H_{5} O H}^{⊖} = 0.06 V$ , Then $E_{c e l l}^{⊖}$ of the reaction taking place in alcohol meter is :

The correct option is B $1. 27 V$
$E_{c e l l}^{⊖} = (E_{C r_{2}}^{⊖} O_{2}^{2 -} | C r^{3 +})_{r e d u c t i o n a t c a t h o d e}$
$- (E_{C H_{3}}^{⊖} C O O H | C_{2} H_{5} O H)_{r e d u c t i o n a t a n o d e}$
= 1.33 - 0.06 = 1.27 V