Give the method to determine the potential of any half-cell.

  • A
    By connecting it to a Standard Hydrogen Electrode $(SHE)$ and measuring the cell potential.
  • B
    By measuring the resistance of the solution.
  • C
    By using a potentiometer directly on the half-cell.
  • D
    By calculating the concentration of ions.

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Similar Questions

$Cr_2O_7^{2-} + I^{-} \to I_2 + Cr^{3+}$
$E^0_{cell} = 0.79 \ V$
$E^0_{Cr_2O_7^{2-}} = 1.33 \ V$,$E^0_{I_2}$ is ........... $V$

The standard electrode potential $(E^{\circ}_{M^{3+}/M^{2+}})$ for $V$,$Cr$,$Mn$,and $Co$ are $-0.26 \ V$,$-0.41 \ V$,$+1.57 \ V$,and $+1.97 \ V$,respectively. The metal ions which can liberate $H_2$ from a dilute acid are

The standard reduction potentials of $Zn^{2+}|Zn$,$Cu^{2+}|Cu$ and $Ag^{+}|Ag$ are respectively $-0.76 \ V$,$0.34 \ V$ and $0.80 \ V$. The following cells were constructed:
$(1)$ $Zn|Zn^{2+}||Cu^{2+}|Cu$
$(2)$ $Zn|Zn^{2+}||Ag^{+}|Ag$
$(3)$ $Cu|Cu^{2+}||Ag^{+}|Ag$
What is the correct order of $E_{\text{cell}}^{\circ}$ of these cells?

Given
$E^o_{\frac{1}{2}Cl_2/Cl^-} = 1.36 \ V$,$E^o_{Cr^{3+}/Cr} = -0.74 \ V$
$E^o_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \ V$,$E^o_{MnO_4^-/Mn^{2+}} = 1.51 \ V$
The correct order of reducing power of the species $(Cr, Cr^{3+}, Mn^{2+}, Cl^-)$ will be:

Among the following metals,the strongest reducing agent is

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