United States presidential election in Virginia, 1884

November 4, 1884

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

Presidential elections
1788-89 1792 1796 1800 1804 1808 1812 1816 1820 1824 1828 1832 1836 1840 1844 1848 1852 1856 1860 1872 1876 1880 1884 1888 1892 1896 1900 1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016
Presidential primaries
Democratic 2008 Republican 2008 2012
U.S. Senate elections
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U.S. House elections
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Special elections
U.S. Senate, 1920 U.S. Senate, 1933 U.S. Senate, 1946 U.S. Senate, 1966 1st congressional district, 2007

Gubernatorial elections
1851 1855 1859 1863 (Confederate) 1863 (Union) 1869 1873 1877 1881 1885 1889 1893 1897 1901 1905 1909 1913 1917 1921 1925 1929 1933 1937 1941 1945 1949 1953 1957 1961 1965 1969 1973 1977 1981 1985 1989 1993 1997 2001 2005 2009 2013 2017
Lieutenant Governor elections
2013 2017
Attorney General elections
2005 2013 2017
House of Delegates elections
2009
Other elections
2007 2009 2011 2013

The 1884 United States presidential election in Virginia took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose twelve representatives, or electors to the Electoral College, who voted for President and Vice President.

Virginia voted for the Democratic candidate, New York Governor Grover Cleveland over the Republican candidate, former Secretary of State James G. Blaine. Cleveland won Virginia by a margin of 2.15%.

Results

United States presidential election in Virginia, 1884^[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Grover Cleveland	145,491	51.05%	12
	Republican	James G. Blaine	139,356	48.90%	0
	Prohibition	John St. John	130	0.05%	0
Totals			284,977	100.0%	12

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