Belgian general election, 1888

Belgian general election, 1888
Belgium
12 June 1888 (1888-06-12)

69 of the 138 seats in the Chamber of Representatives
70 seats needed for a majority
  First party Second party
 
Party Catholic Liberal
Last election 98 seats 40 seats
Seats won 98 40
Seat change Steady0 Steady0

Government before election

Beernaert Government
Catholic

Elected Government

Beernaert Government
Catholic

This article is part of a series on the
politics and government of
Belgium
Constitution
Foreign relations

Partial general elections were held in Belgium on 12 June 1888.[1][2] The result was a victory for the Catholic Party, which won 98 of the 138 seats in the Chamber of Representatives and 47 of the 69 seats in the Senate.[2]

Under the alternating system, elections were held in only five out of the nine provinces: Antwerp, Brabant, Luxembourg, Namur and West Flanders.

Results

Chamber of Representatives

Party Votes % Seats
Won Total +/–
Catholic Party31,27358.466980
Liberal Party19,96737.33400
Others2,2774.300New
Invalid/blank votes
Total53,517100691380
Registered voters/turnout73,276
Source: Mackie & Rose,[3] Sternberger et al

Senate

Party Votes % Seats
Catholic Party 47
Liberal Party 18
Independents 4
Total 69
Source: Sternberger et al.

References

  1. Codebook Constituency-level Elections Archive, 2003
  2. 1 2 Sternberger, D, Vogel, B & Nohlen, D (1969) Die Wahl der Parlamente: Band I: Europa - Erster Halbband, p105
  3. Thomas T Mackie & Richard Rose (1991) The International Almanac of Electoral History, Macmillan, pp50–51
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