the following cell is constructed. beaker 1 contains hydrogen electrode and has solution of $1.0 \mathrm{M} \mathrm{HCL}(\mathrm{aq})$. beaker 2 contains silver and has a solution of $1.0 \mathrm{M}$ $\mathrm{Cu}(\mathrm{NO} 3) 2(\mathrm{aq})$. PREDICT THE READING ON THE VOLTMETER:
OV
$$
\begin{array}{l}
-0.34 \mathrm{~V} \
-0.80 \mathrm{~V} \
+0.80 \mathrm{~V} \
+0.34 \mathrm{~V}
\end{array}
$$

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Accepted Answer

***D*** ∻Key Concept∻ ⚹Standard Electrode Potentials⚹ ∻Explanation∻ ⚹The cell potential is calculated using the standard reduction potentials of the two half-cells. The standard hydrogen electrode (SHE) has a potential of 0 V, and the standard reduction potential for the Cu^2+/Cu half-cell is +0.34 V. Since the Cu^2+/Cu half-cell has a higher reduction potential, it will act as the cathode and the SHE will act as the anode. The cell potential is the difference between the cathode and anode potentials: E_cell = E_cathode - E_anode = +0.34 V - 0 V = +0.34 V.⚹◊What is the standard reduction potential for the hydrogen electrode in the given electrochemical cell?⍭ Generate me a similar question◊

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