For $Cr_2O_7^{2-} + 14H^{+} + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$,$E^0 = 1.33 \ V$. Given $[Cr_2O_7^{2-}] = 4.5 \ mmol$,$[Cr^{3+}] = 1.5 \ mmol$ and $E = 1.067 \ V$,calculate the $pH$ of the solution.

  • A
    $2$
  • B
    $3$
  • C
    $2.5$
  • D
    $1.5$

Explore More

Similar Questions

For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

Difficult
View Solution

What will be the $pH$ of a solution which is in contact with a hydrogen electrode having an oxidation potential of $0.177 \ V$?

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

The standard e.m.f. of a cell,involving one electron change,is found to be $0.591 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction is: $(F = 96,500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

Give the formula to calculate the equilibrium constant $K_C$ of any electrochemical cell.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo