The cell Pt-H2g- 0-1 bar-H-aq- pH - X - Cl-aq- 1 M-Hg2Cl2-Hg-Pt- has e-m-f- of 0-5755 V at 25-C- The SOP of calomel electrode is -0-28 V- then pH of the solution will be-

The correct option is C 5.5 Anode: #160; #160; #160; 1 2 H _ 2 ⟶ H ^ + + e ^ Cathode: #160; 1 2 H g_ 2 Cl _ 2 ⟶ H ^ + +Hg+ Cl ^ ...

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

Faraday's First Law

The cell Pt...

Question

The cell $P t | H_{2} (g, 0.1 b a r) | H^{+} (a q), p H = X | | C l^{-} (a q, 1 M) | H g_{2} C l_{2} | H g | P t$ ,
has e.m.f. of $0.5755 V$ at $25^{\circ} C$ . The SOP of calomel electrode is $- 0.28 V$ , then $p H$ of the solution will be:

11

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

4.5

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

5.5

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

none of these

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

Open in App

Solution

Suggest Corrections

Similar questions

P t, H_{2 (g)} (1 atm) / H^{+}, p H = x | |

0.67 V

25^{\circ} C

- 0.28 V

p H

\frac{2.303 R T}{F} = 0.0591 V

H_{2} (1 a t m) | H^{+} (p H = x) | |

25^{\circ} C

- 0.28 V

^{o}

P t (s) | H_{2} (1 a t m) | H^{+} (? M) | | C l^{-} (1 M) | H g_{2} C l_{2} (s) | H g (l)

E_{C l^{-} / H g_{2} C l_{2} / H g} = 0.28 V

0.67

25

^{o} C

H

0.28 V

0.634

298

P t [H_{2} (1 a t m) | H^{+} (p H = x) ∥ K C l (1 N) | H g_{2} C l_{2} (s) | H g (l)

E^{o}

C l^{-} | H g_{2} C l_{2} | H g

0.28

Join BYJU'S Learning Program