A hydrogen electrode is dipped in a solution at `25^(@)C`. The potential of cell is `-0.177 V`. Calcualte the `pH` of the solution.

Question

A hydrogen electrode is dipped in a solution at `25^(@)C`. The potential of cell is `-0.177 V`. Calcualte the `pH` of the solution.

A hydrogen electrode is dipped in a solution at `25^(@)C`. The potential of cell is `-0.177 V`. Calcualte the `pH` of the solution.
A. 0.177 V
B. `-0.177 V `
C. 0.087 V
D. 0.059 V

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

Normal aluminium electrode coupled with normal hydrogen electrode gives an emf of 1.66 volts . So the standard electrode potential of aluminium is

2 Answers 4 Views

The electrode potential of calomel electrode , used as a reference electrode

2 Answers 4 Views

A standard hydrogen electrode has zero electrode potential is assumed to be

2 Answers 4 Views

For the cell prepared from electrode A and B , Electrode `A : Cr_(2)O_(7)^(2-) | Cr^(3+) , E_("red")^(@) = 1.33 V` and Electrode `B : Fe^(3+) | Fe^(2+

2 Answers 5 Views

Electrode potential of hydrogen electrode is 18 m V then `[H^(+)]` is -

2 Answers 4 Views

If the standard electrode poten tial of `Cu^(2+)//Cu` electrode is 0.34V. What is the electrode potential of 0.01 M concentration of `Cu^(2+)`?

2 Answers 6 Views

A cell when dipped in 0.5 M sucrose solution has no effect but when the same cell will be dipped in 0.5 M NaCl solution the cell will

2 Answers 4 Views

A zinc rod is dipped in 0.1 M `ZnSO_(4)` solution. The salt is 95% dissociated of this dilution at 298 K. Calculate electrode potential. `(E_(Zn^(2+)/

2 Answers 6 Views

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)` M hydrogen ions. The emf of the cell is 0.1

2 Answers 5 Views

Calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode , if the cell potential potential of the cell, `Ni//N^(2+)(0.01 M)//Cu is 0.

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
`E_(RP) = E_(RP)^(@) + (0.0592)/(1) "log"_(10) [H^(+)]`
`= 0 + 0.0592 xx (-3) = -0.177` V

5 views Answered 2023-02-05 15:29