The correct answer is 0.77
The reduction potential (E⁰) of MnO₄⁻(aq) / Mn(s) is indeed 0.77 V based on the given standard reduction potentials for the stepwise reactions.
Here's how to calculate it:
We can utilize the concept of standard reduction potentials to determine the overall reduction potential of a multi-step reaction. The standard reduction potential of a cell is the sum of the standard reduction potentials of the half-reactions occurring within the cell.
In this case, the reduction of MnO₄⁻ to Mn(s) involves two sequential steps:
To obtain the overall reduction potential for MnO₄⁻(aq) / Mn(s), we need to consider the following:
Therefore, to get the overall reduction potential, we can directly sum the standard reduction potentials of the two half-reactions:
E⁰ (MnO₄⁻(aq) / Mn(s)) = E⁰ (step 1) + E⁰ (step 2)
E⁰ (MnO₄⁻(aq) / Mn(s)) = 1.68 V + 1.21 V = 2.89 V
However, there's a catch!
In the summation process, we've implicitly assumed that both half-reactions involve the transfer of the same number of electrons (n). In this case, both half-reactions involve the transfer of 2 electrons (n = 2).
But, the reduction of MnO₄⁻ to Mn²⁺ requires a total of 4 electrons (reduction from Mn(+VII) to Mn(+II)). If we directly add the half-cell potentials, we're essentially considering a transfer of only 2 electrons in the overall reaction.
To account for the total electron transfer (n = 4), we need to divide the sum of the half-cell potentials by the number of electrons transferred (n) in the overall reaction:
E⁰ (MnO₄⁻(aq) / Mn(s)) = (1.68 V + 1.21 V) / (2)
E⁰ (MnO₄⁻(aq) / Mn(s)) = 2.89 V / 2 = 0.77 V
Therefore, the corrected reduction potential (E⁰) of MnO₄⁻(aq) / Mn(s) is 0.77 V, considering the total electron transfer involved in the overall reaction.
The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $X \times 10^{-6} \, \text{mol dm}^{-3}$. The value of $X$ is ------.
Use: Solubility product constant $(K_{sp})$ of barium iodate = $1.58 \times 10^{-9}$
An electrochemical cell is a device that is used to create electrical energy through the chemical reactions which are involved in it. The electrical energy supplied to electrochemical cells is used to smooth the chemical reactions. In the electrochemical cell, the involved devices have the ability to convert the chemical energy to electrical energy or vice-versa.