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?

E = 0.062 V

Explanation:

(a) See the attached file for the answer

(b)

Calculating the voltage (E) using the formula;

E = - (2.303RT/nf)log Cathode/Anode

Where,

R = 8.314 J/K/mol

T = 35°C = 308 K

F- Faraday's constant = 96500 C/mol,

n = number of moles of electron = 2

Substituting, we have

E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)

   = -0.031 * -2

  = 0.062V

Therefore, the voltmeter will show a voltage of 0.062 V

h let me think

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.