Answer:
1. The positive electrode is the one on the left
2. Ecell = 0.0582V
Note: The question is incomplete.
The complete question is given below:
A certain metal M forms a soluble sulfate salt M2SO4. Suppose the left half cell of a galvanic cell apparatus is filled with a 5.00 M solution of M2SO4 and the right half cell with a 500 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 °C. Which electrode will be positive? What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.
Step-by-step explanation:
1. This is a case of a concentration cell. A concentration cell acts to dilute the more concentrated solution and concentrate the more dilute solution, creating a voltage as the cell reaches an equilibrium. This is achieved by transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.
Electrons flow from the right electrode to the left electrode. Therefore, the left electrode is the positive electrode (cathode).
2. See attachment below for solutions