Question textConsider the following electrochemical cell and table of reduction potentials. What is the cell voltage at standard conditions (Eocell)? Give your answer to 2 decimal places, do not include units. Answer 1 Question 7[input] VAccording to the above diagram, in which direction through conducting wire, will electrons flow?Answer 2 Question 7[select: , left to right, Right to left]

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To tackle this problem, first identify the two half-reactions involved in the electrochemical cell from the diagram and the given table of standard reduction potentials. The half-reaction with the higher reduction potential will occur at the cathode (the site of reduction), while the half-reaction with the lower reduction potential will occur at the anode (the site of oxidation).

Option analysis for the cell voltage:
- Consider the potential values for the two electrodes shown in the diagram. If we take Fe2+/Fe as one half-reaction with E°red = -0.44 V and Pb2+/Pb as the other with E°red = -0.13 V, the Fe couple has the more negative (lower) reduction potential, so Fe is the anode and Pb is the cathode in this configuration.
- The standard cell potential is then E°cell = E°red(cathode) − E°red(anode) = (-0.13 V) − (-0.44 V) = 0.31 V. This matches the provided answer value of 0.31 (to two decimals, no units). Note that the sign convention used here corresponds to a spontaneous galvanic cell with a positive E°cell.

Direction of electron flow through the conducting wire:
- In an electrochemical cell, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs).
- Based on the chosen assignments, the anode is the Fe electrode (Fe2+/Fe with E°red = -0.44 V), and the cathode is the Pb electrode (Pb2+/Pb with E°red = -0.13 V).
- Therefore, electrons will flow from the Fe electrode to the Pb electrode through the conducting wire.
- Translating this to the diagram’s left-right labeling, electrons flow from the right-hand Fe electrode to the left-hand Pb electrode, i.e., Right to left.

Summary of reasoning steps:
- Compare E°red values to assign cathode and anode: Pb2+/Pb (-0.13 V) is cathode; Fe2+/Fe (-0.44 V) is anode.
- Compute E°cell = (-0.13) − (-0.44) = 0.31 V.
- Apply the rule that electrons move from anode to cathode in a galvanic cell, which gives Right to left for the provided diagram.

If any of these positional interpretations of the diagram were different, you would re-identify which half-reaction sits at the cathode (the one with the higher E°red) and recalculate E°cell using the same formula, ensuring the result is positive for a spontaneous cell.

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