A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.00 M solution of MSO, and the ight half cell with a 30.0 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 35.0 °C lf right 410 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.

Explanation:

It is known that for high concentration of , reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, = 0 and the general reaction equation is as follows.

                3.00 M        n = 2       30 mM

         E =

            =

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2