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For the electrochemical cell M | M²+ || X²- | X, the standard reduction potentials are: E_red(M²+/M) = 0.46 V and E_red(X/X²-) = 0.34 V (note: E_red(X/X²-) means X gains electrons to form X²-, but given the notation X|X²- at the cathode suggests X is reduced). Given the cell notation M|M²+||X|X²- where M is the anode and X side is the cathode, and E_cell = E_cathode - E_anode. Which statement is correct?

1. E_cell = -0.80 V
2. The reaction M + X -> M²+ + X²- is spontaneous
3. The reaction M²+ + X²- -> M + X is spontaneous
4. E_cell = 0.80 V

Correct answer: The reaction M²+ + X²- -> M + X is spontaneous

Solution

The cell notation M|M²+||X|X²- implies M is the anode (oxidized: M -> M²+ + 2e-) and the right electrode involves X/X²-. Given E_red(M²+/M) = 0.46 V and E_red(X/X²-) = 0.34 V, the species with higher reduction potential (M²+) acts as the cathode (is reduced). Therefore, the actual spontaneous cell has M²+ + X²- -> M + X with E_cell = 0.46 - 0.34 = +0.12 V > 0, confirming this reaction is spontaneous. The cell as written M|M²+||X|X²- is non-spontaneous (E_cell = 0.34 - 0.46 = -0.12 V), but among the options the correct statement is that M²+ + X²- -> M + X is spontaneous.

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